jrbraves26 0 Report post Posted March 15, 2013 You have 50 coins that all look exactly the same. One of them is fake and weighs a few grams more than the others. Given a balance scale (not a triple beam scale), what is the fewest number of weightings you could use to find the fake coin? Explain. Share this post Link to post Share on other sites

0 Pankaj Varma 2 Report post Posted March 15, 2013 (edited) You need 3 or 4 weighings Divide into 3 heaps - 17, 17, and 16 and weigh the first two. You now know which heap has the fake coin. Repeat process with the selected heap - 6,6,5 or 5,5, 6 as the case may be, and weigh the first two. Repeat again 2,2,2 or 2,2,1. If the single coin is fake, no further weighing is required. Else You have to do a 4th weighing if the fake coin is in a group of 2. Edited March 15, 2013 by bonanova Add spoiler Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted March 15, 2013 You have 50 coins that all look exactly the same. One of them is fake and weighs a few grams more than the others. Given a balance scale (not a triple beam scale), what is the fewest number of weightings you could use to find the fake coin? Explain. You have 50 possibilities Each weighing distinguishes among 3 cases. 3x3x3 = 27, so you'd need a fourth weighing for this case. Share this post Link to post Share on other sites

You have 50 coins that all look exactly the same. One of them is fake and weighs a few grams more than the others. Given a balance scale (not a triple beam scale), what is the fewest number of weightings you could use to find the fake coin? Explain.

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